| 1. |
(a) Express the decimal number
-0.187510 in the floating point format M(6) E(4), with both mantissa and
exponent in 2's complement. |
[4] |
|
| 0.1875 |
= |
0.00112 |
|
= |
0.11000 * 2-2 |
| -0.1875 |
= |
101000 * 2-2 |
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| 101000 |
1110 |
| mantissa |
exponent |
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|
One mark for
initial conversion to binary, one mark for moving point, one mark for correct mantissa,
one mark for correct exponent. |
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[4
marks] |
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(b) Perform the following addition
: |
|
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1000 1001BCD + 0111
11012 |
[3] |
|
Convert the answer into hexadecimal
representation. |
[1] |
|
1000 1001BCD |
|
|
= 8910 |
|
|
= 10110012 |
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One mark for
each correct conversion, to a total of two marks |
|
|
 |
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One mark for a
correct addition |
|
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= D616 |
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One mark |
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[4
marks] |
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(c) Derive a simplified POS
(product of sums) expression from the following K map: |
[4] |
|
 |
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The K-map
should be |
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 |
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Two marks for
the above answer; one mark if another grouping is provided which includes more don't cares
but no 1's |
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 |
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Two marks for
the above answer; one mark for correct answer based on wrong grouping; one mark for wrong
answer but correct with respect to grouping given - that is, a consequential error. |
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If candidate
takes SOP route, and correctly derives the POS as the complement of the expression
obtained by grouping the 1's, then they should receive full marks. |
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[4
marks] |
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(d) Give the truth table for the
block diagram shown below: |
[4] |
|
 |
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One mark for
each correct row; half marks would be unnecessary here. |
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[4
marks] |
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(e) Consider the circuit shown
below |
[4] |
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 |
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and complete the waveform for
output Q, as input A varies as shown over four clock cycles: |
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 |
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The
waveform for Q should be as follows |
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One mark for
each correct clock cycle: a, b, c, and d. |
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[4
marks] |
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